In the derivation of capillary rise, we take excess pressure due to surface tension in upward direction as equal to $2T/r$.

Can some one please explain how does this come about?

  • $\begingroup$ Refer to Young-Laplace equation. The answers given are correct but in think this is more fundamental $\endgroup$ – Ishan Jawale Jun 15 '19 at 13:05

The derivation can be thought of as this:


  • Radius of Capillary be $r$

  • Density of the liquid $\rho$

  • Height of the liquid be $h$

  • Surface Tension of Liquid be $T$

  • Contact angle $\theta$

Weight of liquid inside capillary = Volume * Density * $g$ $$=\pi r^2 h \rho g$$Which is the downward force, and the force that is balancing this is the force due to Surface tension.

Now, Surface tension is defined as the Force acting on a line which is on the surface. In this case the surface tension is acting on the circumference.

Hence, total force upwards: component of Surface Tension upwards * length of the line it acts on $$T\cos\theta (2\pi r)$$ The $\sin\theta$ components gets cancelled as it is radially outward throughout the circumference.

Equating the forces we get: $$\pi r^2 h \rho g = T\cos\theta (2\pi r)$$ $$\implies h = \frac{2T\cos\theta}{r\rho g}$$

Note: In cases of some liquids the $\theta$ is very close to $0$ degrees and hence the cos$\theta$ term can be taken as $1$.

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We know that $p=dgh$, where $d=$ density, $g=$ acceleration due to gravity, $h=$ height. Therefore, from the above equation $p=2T/r$ where $\cos \theta =1$.

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